# This problem can be solved with some math based on pascal's triangle, but
# this solution with some good optimizations is fast enough. 
# Runs "instantly" on a modern system
from math import factorial

# gives the number of possibilities for r for a given n in nCr
def num_configs(n):
	top = factorial(n)
	r=int(n/2) # nCr is largest at r=n/2. Then recudes in either direction.
	denom = factorial(r)*factorial(n-r)
	cnt=0
	while top/denom > 1000000 and r >0:
		r-=1
		cnt+=1 if cnt == 0 and int(n/2)==n/2 else 2
		denom = factorial(r)*factorial(n-r)
	return cnt

total=0
for i in range(1,101):
	total+= num_configs(i)

print(total)